The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

The standard deviation of the sample is the same as the standard error of the mean. So

$SE_{M} = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\sigma = 0.35, n = 20$

$SE_{M} = \frac{\sigma}{\sqrt{n}}$

$SE_{M} = \frac{0.35}{\sqrt{20}}$

$SE_{M} = 0.0783$

The standard error of the mean is 0.0783.

7 0

3 years ago

PLEASE HELP ASAP!!!

barxatty [35]

Answers:

<u>24000 dollars</u> invested at 4%
<u>18000 dollars</u> was invested at 7%

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Work Shown:

x = amount invested at 4%

If she invests x dollars at 4%, then the rest (42000-x) must be invested at the other rate of 7%

She earns 0.04x dollars from that first account and 0.07(42000-x) dollars from the second account

This means we have

0.04x+0.07(42000-x)

0.04x+0.07*42000-0.07x

0.04x+2940-0.07x

-0.03x+2940

This represents the total amount of money earned after 1 year.

We're told the amount earned in interest is $2220, so we can say,

-0.03x+2940 = 2220

-0.03x = 2220-2940

-0.03x = -720

x = -720/(-0.03)

x = 24000 dollars is the amount invested at 4%

42000-x = 42000-24000 = 18000 dollars was invested at 7%

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As a check, we can see that

18000+24000 = 42000

and also

0.04x = 0.04*24000 = 960 earned from the first account

0.07*18000 = 1260 earned from the second account

1260+960 = 2220 is the total interest earned from both accounts combined

This confirms our answers.

7 0

3 years ago